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Conditioned on the parameters, the data augmentation labels are independent. By Bayes' theorem, the conditional probability of component $i$ for observation $j$ isThis is the mixture responsibility. Draw an independent uniform random variable and select $i$ using the cumulative sums of $(r_{j1},\dots ,r_{jk})$. Refresh every label before recomputing the component sufficient statistics for the next Gibbs sampler sweep. Numerically, form the log scores $\log {\omega }_i-\frac{1}{2}\log v_i-(x_j-{\mu }_i{)}^2/(2v_i)$, subtract their maximum and exponentiate before normalization.

For events with $\mathbb{P}(B)>0$, the conditional probability isUse the sample space $\Omega =\{1,2,3,4\}\times \{0,1{\}}^4$, where the first coordinate is the chosen coin and the other coordinates record which of its two physically distinguished faces lands upwards on each toss. There are $4\cdot 16=64$ equally likely elementary outcomes. Label the double-headed coin by one. For that coin both face indices mean heads; for each other coin only face index one means heads. In particular, strings of observed heads and tails are not equally likely outcomes.

Let $E$ be the event that the first three tosses show heads. With coin one all sixteen face-index strings are permitted. For each ordinary coin the first three indices must be one and the fourth is free, giving two strings. Thus $|E|=16+3\cdot 2=22$. Of these outcomes, sixteen from the double-headed coin and one from each ordinary coin also have a fourth head. HenceEquivalently, Bayes' theorem gives conditional probability $8/11$ of having selected the double-headed coin after $E$; the fourth-head probability is then $8/11+(3/11)(1/2)=19/22$. The tosses are independent conditional on the selected coin, while mixing over the same coin explains their unconditional dependence.

Because the events are pairwise disjoint events, their count can only be zero or one. The event that the count is zero is the complement of an event of their union. Finite additivity therefore gives

Now use $N\ge 1$ for the fixed number of independently located ships. Normalize the radius to one and write their positions as unit vectors $a_1,\dots ,a_N$. A point $x$ on the sphere is reached by ship $i$ precisely when $a_i\cdot x>0$, so each ship reaches an open hemisphere. Coverage fails exactly when there is a unit vector $x$ with $a_i\cdot x\le 0$ for every $i$. Equivalently, all ship positions lie in some closed hemisphere.

With probability one no pair of ship vectors is antipodal and no three are linearly dependent: the exceptional choices lie on sets of zero surface area. For such a configuration, containment in a closed hemisphere implies containment in an open hemisphere. To see this, choose its inward normal $u$, so $u\cdot a_i\ge 0$. At most two vectors lie on its boundary. If there are two, their sum has positive dot product with each, since they are not antipodal; if there is one, use that vector itself. Perturb $u$ slightly in this direction. The boundary dot products become positive while all already positive dot products remain positive. Thus the strict and non-strict containment conditions have the same probability here.

Condition on the unoriented axes $\{a_i,-a_i\}$. Choose representatives $b_i$ by any fixed sign convention. By uniform sampling and independence, the actual positions are $a_i={\epsilon }_i{b}_i$ with $N$ independent fair signs ${\epsilon }_i\in \{-1,1\}$. Conditional on these axes, each of the $2^N$ sign choices has probability $2^{-N}$.

The great circles $b_i\cdot u=0$ partition the sphere into $R_N=N(N-1)+2$ open regions. In each region the sign vector $\sigma (u)=(\mathrm{sgn}(b_i\cdot u){)}_i$ is constant. No sign vector labels two different regions: its strict inequalities define a convex cone, and normalized line segments within that cone connect any two points of its spherical section. The regions of a general-position great-circle arrangement therefore yield exactly $R_N$ realized sign vectors.

The chosen positions lie in an open hemisphere if and only if some $u$ has $u\cdot a_i>0$ for every $i$, that is, ${\epsilon }_i=\mathrm{sgn}(b_i\cdot u)$ for every $i$. Failure of coverage thus corresponds exactly to one of the $R_N$ realized sign choices. These choices are mutually exclusive events, each of conditional probability $2^{-N}$, so the first part applies:The result is independent of the axes. Averaging gives the coverage of a sphere by random open hemispheres:For $N=1,2,3$ this is zero; for $N=4$ it is $1/8$. If no ship lands, coverage has probability zero separately. The counting uses equiprobable orientation signs, not an assumption that the spherical regions have equal area.

By the definition of conditional probability,ThereforeIn particular $d\log F/dt=-h$ and $F(t)=e^{-H(t)}$ when $F(0)=1$.

Normalize the probability density function using $u=t^c$:The corresponding Weibull distribution has survival functionSince the distribution is continuous, the choices $>$ and $\ge$ have the same probabilities. The conditional probability is consequentlyFor $s,t>0$ and $c>1$, the supplied inequality, applied to $s/t$, gives $(s+t{)}^c>t^c+s^c$. Taking exponentials yieldsThis expresses the increasing hazard function $h_c(t)=c{t}^{c-1}$: surviving to a later age shortens the remaining lifetime in this conditional distribution. At $c=1$ equality holds, as expected from the memoryless property of the exponential distribution.

The general, or asymmetric Lovász local lemma, is as follows. Let $E_1,\dots ,E_m$ be events, with a dependency graph of events having neighbour sets $N(i)$. Each $E_i$ must be independent of the sigma-algebra generated by all the events indexed by its nonneighbours. If numbers $0\le x_i<1$ satisfythenFor completeness, the key inductive bound is $\mathbb{P}(E_i\mid {\bigcap }_{j\in S}{E}_j^c)\le x_i$ whenever $i\notin S$. Split $S$ into neighbours $S_1$ of $i$ and nonneighbours $S_2$. If $S_1$ is empty, the dependency condition gives the bound directly. Otherwise,The denominator bound follows by exposing the exclusions in $S_1$ one at a time and using the induction hypothesis with strictly fewer conditioning events. The same induction ensures that all conditioning events have positive probability. Finally multiply the successive conditional probabilities of avoiding $E_1,\dots ,E_m$ to obtain the stated lower bound.

For the symmetric form, suppose each event has probability at most $q$ and the dependency graph has maximum degree $D\ge 1$. Choose $x_i=1/(D+1)$. ThenThe final inequality follows from $(1+1/D{)}^D\le e$. Thus $eq(D+1)\le 1$ suffices for positive probability that no bad event occurs. If $D=0$, the dependency condition gives joint independence and the same criterion implies $q<1$, so the conclusion follows directly.

Apply this to a fair independent two-colouring of the hypergraph vertices, with $E_W$ the event that hypergraph edge $W$ is monochromatic. Its probability is $q=2^{1-r}$. Events on disjoint hypergraph edges use disjoint independent variables, giving a dependency graph by joining intersecting hypergraph edges. Each hypergraph edge meets at mostother hypergraph edges; repeated counting only enlarges this upper bound. The slightly looser bound $D+1\le r(\Delta +1)$ is enough for the printed condition. Indeed,The symmetric Lovász local lemma therefore gives a proper two-colouring. A hypergraph with no hypergraph edges is trivially colourable and needs no dependency calculation.

The statement is true. Put $C_k={\bigcap }_{r=1}^k{A}_r$. The assumed $P(C_{n-1})>0$ ensures that every conditioning event $C_{k-1}$, $2\le k\le n$, has positive probability. By the definition of conditional probability,Multiplying makes the fractions telescope, givingThis proves the chain rule for probabilities here without any assumption of independence. The final numerator $P(C_n)$ may be zero; all denominators remain positive.

Let $o(\omega )$ be the number of open edges, $k(\omega )$ the number of connected components in the open subgraph, counting isolated vertices, and $m=|E|$. Summing the joint weight over spins requires a constant spin on every open component. There are exactly $q^{k(\omega )}$ choices. Thus the edge marginal isThis is the random-cluster measure, and the joint normalizing constant is $Z_{\mathrm{RC}}$.

Summing instead over edge configurations factors independently over edges. For a fixed spin configuration the resulting weight issince $1-p=e^{-\beta }$. The factor $e^{-\beta m}$ is independent of the spins. Hence the spin marginal is exactly the ferromagnetic Potts measure in the conventionIts partition function satisfies $Z_{\mathrm{RC}}=e^{-\beta m}{Z}_{\mathrm{P}}$.

Conditionally on $\sigma$, edge states are independent: an edge with different endpoint spins is closed with probability one, while one with equal endpoint spins is open with probability $p$. Conditionally on $\omega$, color its $k(\omega )$ open components independently and uniformly from $\{1,\dots ,q\}$. Equivalently, every compatible spin configuration has conditional probability $q^{-k(\omega )}$, and all incompatible configurations have probability zero. These are the two conditional laws of the Edwards-Sokal coupling.

If $x,y$ belong to the same open component, their colors agree surely; otherwise the two independent component colors agree with probability $1/q$. ThereforeRearranging proves the spin-connectivity identity for the Potts model:

Phase transitions refer to infinite-volume limits, not a single finite graph. The identity shows that connectivity is exactly the excess spin agreement above its independent-color baseline. More directly, use a wired random-cluster boundary condition and fix the corresponding boundary spin to color $1$. Its component has color $1$, while every finite component has a uniform color. In the compatible infinite-volume coupling,Consequently normalized spontaneous magnetization isThe onset of a wired infinite cluster and the onset of spontaneous magnetization in the Potts model are the same transition under $p=1-e^{-\beta }$, giving ${\beta }_c(q)=-\log (1-p_c(q))$. This also transfers any jump or continuity of that order parameter between the two descriptions; it makes no assumption about behavior exactly at the critical point.

Choose an orthonormal eigenbasis of $V$, with eigenphases ${\varphi }_0,{\varphi }_1$, counting multiplicity. The maximally mixed state is basis independent, soLinearity means that the experiment is the equal mixture of the two eigenstate experiments. After the inverse quantum Fourier transform, the joint state is $\frac{1}{2}{\sum }_{j=0}^1|{\chi }_j⟩⟨{\chi }_j|\otimes |v_j⟩⟨v_j|$, where $|{\chi }_j⟩={\sum }_m{a}_m({\varphi }_j)|m⟩$ andThus the exact output distribution for quantum phase estimation with a maximally mixed target isFor a nonzero phase mismatch, the squared probability amplitude is ${\sin }^2[L(\varphi -2\pi m/L)/2]/\{L^2{\sin }^2[(\varphi -2\pi m/L)/2]\}$, with value one at an exact match. If both eigenphases are exactly representable on the $L$-point grid, their two labels occur with probability $1/2$ each; coincident labels have their probabilities added. For general eigenphases there can be nonzero probabilities at other labels, so it would not be exact to list only the two best approximations with probability $1/2$ each. Each eigenphase is sampled with spectral weight $1/2$, and its nearest-label event has conditional probability at least $4/{\pi }^2$. In particular a nearest label for one eigenphase has unconditional probability at least $2/{\pi }^2$, with additional contributions possible from the other phase.

The law of total probability and the definition of conditional probability give Bayes' theorem:The denominator is positive because $P(E)>0$.

For the observed cargo, set $p_A=0.05$, $p_B=0.03$, $p_C=0.01$. The binomial likelihood under origin $i$ isEqual prior probabilities and Bayes' theorem make the posterior proportional to these likelihoods. The binomial coefficient cancels in the likelihood ratios:Thus $L_A/L_B\approx 4.99\times 10^{-45}$ and $L_C/L_B\approx 2.74\times 10^{-9}$, givingEven the permitted approximation $\log (1-p)\approx -p$ gives log ratios about $-93.83$ and $-23.72$, which already show overwhelming posterior preference for B. The exact logarithms are preferable because multiplication by $9800$ magnifies the approximation error. An observed contamination fraction halfway between two hypothesized rates does not give equal likelihoods: the binomial distribution assigns asymmetric probabilities to those deviations, especially at this large sample size.

The events $B\cap A_i$ are pairwise disjoint and their union is $B$. Finite additivity and the definition of conditional probability therefore give the law of total probability:For the birthday problem, let $D$ mean that all birthdays differ. For $n\le 365$, sequential conditioning on previous distinct birthdays and independence of random variables giveHere each factor uses the stated bound $1-x\le e^{-x}$. For $n\ge 29$, $n(n-1)\ge 812>730\log 3$, soFor $n>365$ a match is certain by the pigeonhole principle. The threshold from the exponential bound is $(1+\sqrt{1+2920\log 3})/2\approx 28.8$, so $29$ suffices.

Write the Cox proportional-hazards model as $h_i(t)=h_0(t)u_i$, where $u_i=\exp ({\beta }^T{z}_i)$ is the hazard multiplier and $h_0$ is the unspecified baseline hazard. At an untied event time $t_j$, let $R_j$ be the risk set immediately before the event and $i_j$ its failing individual. With independent censoring conditional on the covariates, the conditional probability that individual $i$ supplies the next event, given an event at that time and the risk set, is obtained by dividing its instantaneous event rate by the aggregate rate:More precisely this is the limit of the conditional probability for one event in a short interval; simultaneous events have smaller-order probability in the continuous model. Multiplication of these successive conditional contributions gives the Cox partial likelihoodCensored individuals remain in each risk set until their censoring time but supply no numerator. Conditioning removes the baseline hazard; this is a partial likelihood for $\beta$, rather than a complete likelihood for event and censoring times.

Ties require an observation model or an approximation. Suppose a recorded tie hides the order of $d$ continuous events, with no intervening entry or censoring. Put $D$ for the tied event set and $S={\sum }_{i\in R}{u}_i$. An exact marginal contribution for the coarsened ranks is the sum of the ordinary rank contributions over every possible order:For example, if $A,B$ fail together while $C$ remains at risk, the two possible orders giveThe risk set is depleted after the first of the two events. This expression concerns the missing event order; it is not the full likelihood of an interval in which the events occurred.

Convenient alternatives are the Breslow approximation for tied event times and the Efron approximation for tied event times:The first keeps the initial denominator at every event; the second removes an average share of the tied-event multipliers successively. In the example their denominators are $S^2$ and $S[S-(u_A+u_B)/2]$. These conventional partial-likelihood factors omit multiplicities independent of $\beta$, which do not affect its estimate; they should not be mistaken for normalized probabilities of the unordered event set.

For genuinely discrete event times there is another exact construction. If individual event odds are proportional to $u_i$, conditioning on exactly $d$ events gives the exact tied-set conditional likelihoodFor the same three individuals this is $u_A{u}_B/(u_A{u}_B+u_A{u}_C+u_B{u}_C)$. It is not generally equal to the continuous-time rank sum: the assumptions differ.

Large tie groups make naive summation over $d!$ orders or $\left(\genfrac{}{}{0ex}{}{|R|}{d}\right)$ subsets impractical. Recursion helps with the subset denominator, and Breslow or Efron avoids exhaustive enumeration. However, extensive ties can also indicate that time is measured too coarsely for an exact continuous-time ordering to be a useful target.

For example, if patients are assessed only at annual visits, use one row per patient-year still at risk and model the probability $p_{ij}$ of an event in that interval. A grouped proportional-hazards model follows directly by integrating the hazard over the interval, with covariates constant there:Fit the resulting person-period likelihood, with Bernoulli distributions for the interval event indicators and interval-specific intercepts. Many events in the same year are then ordinary observations rather than a combinatorial ordering problem. This requires a compatible interval-observation scheme and appropriate independent censoring; it avoids pretending that unobserved within-year event times are known.

The event that $H_1$ is hit but $H_2$ is not has probability $1/3-1/12=1/4$. Dividing by the probability $1/3$ that $H_1$ is covered, using the definition of conditional probability, givesThis conditioning concerns angular coverage of a sector; the two house-hit events are not independent.

For coverage of antipodal points by a randomly oriented sector, the house bearings differ by $\pi$. For $\varphi <\pi$, no single sector can cover both. For $\varphi \ge \pi$, the two allowable arcs of centre bearings overlap in two components, each of length $\varphi -\pi$. Thus the conditional probability of a double hit is $(\varphi -\pi )/\pi$, not $(\varphi -\pi )/(2\pi )$. Averaging against the probability density function givesPut $v=\varphi -\pi$ to obtain $\frac{1}{2{\pi }^3}{\int }_0^{\pi }v(\pi -v)dv$. Therefore

Conditional on $\Phi =\varphi$, the uniform distribution of the sector's centre angle gives a range of possible centre bearings of length $\varphi$ that hit a specified house. Independence of the two angles therefore gives the conditional probability $\varphi /(2\pi )$. By the law of total expectation,

By the definition of conditional probability,Since the $B_j$ form a disjoint partition of the sample space, the law of total probability gives $\mathbb{P}(A)={\sum }_j\mathbb{P}(A\mid B_j)\mathbb{P}(B_j)$. Substitution proves Bayes' theorem in the required form:The hypotheses make the denominator positive and each conditioning probability well defined.

The independent uniform distributions give a constant joint probability density $1$ on the unit square. For $0<x<1$, integrate over $x<u<v^2$:For the random quadratic with uniform coefficients, real roots require its discriminant to be nonnegative, equivalently $U\le V^2$. The boundary has probability zero, and the area under this parabola isOn that event the roots are $-V\pm \sqrt{V^2-U}$, both nonpositive. Bounding both absolute values by one therefore amounts to bounding the more negative root:Since $0\le V\le 1$, squaring the inequality is legitimate and gives $U\ge 2V-1$. The admissible unit-square region is $\max (0,2v-1)\le u\le v^2$. Its probability isDividing by the real-root probability gives the requested conditional probability:

For events with $P(B)>0$, the conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$. Applying the definition in both orders gives Bayes' theorem:Conditionally on $N=n$, independent fair tosses give a binomial distribution for the head count, so $P(H=1\mid N=n)=n{2}^{-n}$. The joint probability mass function is consequentlyUse the law of total probability and the differentiated geometric series, ${\sum }_{n\ge 1}n{r}^n=r/(1-r{)}^2$, to obtain $P(H=1)=4/9$. The posterior coin count after exactly one head is thereforeThese probabilities sum to one by the same series identity. The conditioning weights possible coin counts by their likelihood of producing exactly one head.

Squaring a variable with the standard normal distribution gives the required chi-squared distribution. Its two inverse images at $y>0$ are $\pm \sqrt{y}$, so the change-of-variable formula givesDraw the branch-selection uniform variable independently of the randomness used for the normal draw. Then the pair $(Y,U)$ has the product law: conditioning on $Y$ leaves $U$ uniform. This independence is exactly what justifies interpreting the subsequent threshold as the conditional probability of selecting the smaller root.

For $x>\mu$, the corresponding smaller root is ${\mu }^2/x$. The conditional probability of selecting the larger root isIt is the same weight as a function of the actual selected value $x$. HenceCombining the two open intervals proves the output has the Inverse Gaussian distribution $IG(\mu ,\lambda )$. The event $Y=0$ has probability zero, so no atom occurs at $\mu$; the limiting density there is finite. The algorithm always selects one of its positive roots, so these branch densities exhaust its probability distribution.